A particle moves along the $x$ -axis. The function $x(t)$ gives the particle's position at any time $t\geq 0$ $x(t)=t^3-3t^2+7t-6$ What is the particle's acceleration $a(t)$ at $t=3$ ? $a(3)=$
Solution: We have a function for the particle's position, and we need to find the particle's acceleration. Acceleration is the rate of change of velocity, which is the rate of change of position. So we need to find the second derivative of $x(t)$. In other words, if $v(t)$ gives the particle's velocity and $a(t)$ gives the particle's acceleration at any time $t\geq 0$, then $a(t)=v'(t)=x''(t)$. Let's differentiate $x(t)$ to find $v(t)$ : $\begin{aligned} v(t)&=x'(t) \\\\ &=\dfrac{d}{dx}[t^3-3t^2+7t-6] \\\\ &=3t^2-6t+7 \end{aligned}$ Let's differentiate $v(t)$ to find $a(t)$ : $\begin{aligned} a(t)&=v'(t) \\\\ &=\dfrac{d}{dt}[3t^2-6t+7] \\\\ &=6t-6 \end{aligned}$ To find the particle's acceleration at $t=3$, we need to evaluate $a(3)$. $\begin{aligned} a({3})&=6({3})-6 \\\\ &=12 \end{aligned}$ In conclusion, the particle's acceleration at $t=3$ is $12$.